2

I know the answer is $\pi$ there is a proof here. Now looking to my textbook (textbook image) the result should be $0$. Using the last equation on the right hand page we have: $$ i\pi(\sin^2(x))'|_{x=0} = 0 $$ and there are no other poles except $0$ so the result should be $0$.

What is the problem? Is the definition for principal value different in the two cases?

Community
  • 1
Jack996
  • 23
  • 1
    $\sin^2(x) / x^2$ is analytic and integrable on $\mathbb{R}$.. so it is a normal Riemann integral, not a principal value – reuns Jun 04 '16 at 08:02

3 Answers3

2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} \color{#f00}{\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x} & = \int_{-\infty}^{\infty}\ \overbrace{\bracks{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k}} ^{\ds{{\sin\pars{x} \over x}}}\ \overbrace{\bracks{\half\int_{-1}^{1}\expo{-\ic qx}\,\dd q}\,\dd x} ^{\ds{{\sin\pars{x} \over x}}} \\[3mm] & = {\pi \over 2}\int_{-1}^{1}\int_{-1}^{1}\ \overbrace{% \int_{-\infty}^{\infty}\expo{\ic\pars{k - q}x}\,{\dd x \over 2\pi}} ^{\ds{\delta\pars{k - q}}}\ \,\dd k\,\dd q = {\pi \over 2}\int_{-1}^{1}\ \overbrace{\int_{-1}^{1}\delta\pars{k - q}\,\dd k}^{\ds{=\ 1}}\ \,\dd q \\[3mm] & = {\pi \over 2}\int_{-1}^{1}\dd q = \color{#f00}{\pi} \end{align}

Felix Marin
  • 89,464
1

That kind of integrals are best managed through integration by parts. We have:

$$ \int_{-\infty}^{+\infty}\left(\frac{\sin x}{x}\right)^2\,dx = \int_{-\infty}^{+\infty}\frac{\frac{d}{dx}\sin^2 x}{x}\,dx = \int_{-\infty}^{+\infty}\frac{\sin(2x)}{x}\,dx = \int_{-\infty}^{+\infty}\frac{\sin z}{z}\,dz=\color{red}{\pi}.$$

Jack D'Aurizio
  • 353,855
0

One may observe that, by the Taylor series expansion, as $x \to 0$, $$ \sin x=x+O(x^3) $$ giving $$ \frac{\sin^2 x}{x^2}=1+O(x^2), $$ on the other hand, as $x \to \infty$, $$ \left|\frac{\sin^2 x}{x^2}\right|\leq \frac1{x^2} $$ consequently the given integral is convergent: $$ 0<\int_{-\infty}^{+\infty}\frac{\sin^2 x}{x^2}\:dx<\infty. $$ No need to consider a principal value: there is no singularity of $\dfrac{\sin^2 x}{x^2}$ over $\mathbb{R}$.

We have

$$ \int_{-\infty}^{+\infty}\frac{\sin^2 x}{x^2}\:dx=2\int_0^{+\infty}\frac{\sin^2 x}{x^2}\:dx=2\cdot \frac{\pi}2=\pi, $$

a proof of the latter integral evaluation may be found here.

Community
  • 1
Olivier Oloa
  • 120,989
  • Right by what is wrong with the equation in my textbook? It seems to be very general. @Olivier – Jack996 Jun 04 '16 at 08:10
  • @Jack996 I think the problem is that in the book when writing $\frac{f(x)}{(x-x_0)^2}$ it clear that the singularity at $x_0$ can't be removed. But in your example $\frac{\sin^2 x}{(x-x_0)^2}=1+O(x^2)$ the singularity has been removed, there is no pole of the integrand anymore... It is not the same situation... Do you see what I mean? – Olivier Oloa Jun 04 '16 at 08:18
  • I see but I didn't quite understand how you proved the integral is convergent. @Olivier – Jack996 Jun 04 '16 at 08:22
  • The given integral is convergent because we may write $\int_{-\infty}^\infty=\int_{-\infty}^{-1}+\int_{-1}^{1}+\int_{1}^{\infty}$. the integral in the middle is convergent since the integrand is continuous over $[-1,1]$ (see taylor's expansion of $\sin$), the other two integrals are convergent since each one is $\leq \int_1^{\infty} \frac1{x^2}dx<\infty$ (we have $|\sin^2 x|\leq1$). – Olivier Oloa Jun 04 '16 at 08:34